一、去除List中重複的String

public List<String> removeStringListDupli(List<String> stringList) {
    Set<String> set = new LinkedHashSet<>();
    set.addAll(stringList);

    stringList.clear();

    stringList.addAll(set);
    return stringList;
}

或使用Java8的寫法：

List<String> unique = list.stream()
    .distinct()
    .collect(Collectors.toList());
 

二、List中物件去重

比如現在有一個 Person類：

    public class Person {
    private Long id;

    private String name;

    public Person(Long id, String name) {
        this.id = id;
        this.name = name;
    }
    //getter&setter...

    @Override
    public String toString() {
        return "Person{" +
                "id=" + id +
                ", name='" + name + '\'' +
                '}';
    }
}
 

重寫Person物件的equals()方法和hashCode()方法：

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Person person = (Person) o;

        if (!id.equals(person.id)) return false;
        return name.equals(person.name);
    }

    @Override
    public int hashCode() {
        int result = id.hashCode();
        result = 31 * result + name.hashCode();
        return result;
    }
 

下面物件去重的程式碼：

        Person p1 = new Person(1l, "jack");
        Person p2 = new Person(3l, "jack chou");
        Person p3 = new Person(2l, "tom");
        Person p4 = new Person(4l, "hanson");
        Person p5 = new Person(5l, "膠布蟲");

        List<Person> persons = Arrays.asList(p1, p2, p3, p4, p5, p5, p1, p2, p2);

        List<Person> personList = new ArrayList<>();
        // 去重
        persons.stream().forEach(
                p -> {
                    if (!personList.contains(p)) {
                        personList.add(p);
                    }
                }
        );
        System.out.println(personList);

List 的contains()方法底層實現使用物件的equals方法去比較的，其實重寫equals()就好，但重寫了equals最好將hashCode也重寫了。

可以參見:

http://stackoverflow.com/questions/30745048/how-to-remove-duplicate-objects-from-java-arraylist 

http://blog.csdn.net/growing_tree/article/details/46622579

三、根據物件的屬性去重

下面要根據Person物件的id去重，那該怎麼做呢？ 
寫一個方法吧:

  public static List<Person> removeDupliById(List<Person> persons) {
        Set<Person> personSet = new TreeSet<>((o1, o2) -> o1.getId().compareTo(o2.getId()));
        personSet.addAll(persons);

        return new ArrayList<>(personSet);
    }

通過Comparator比較器，比較物件屬性，相同就返回0，達到過濾的目的。

再來看比較炫酷的Java8寫法:

import static java.util.Comparator.comparingLong;
import static java.util.stream.Collectors.collectingAndThen;
import static java.util.stream.Collectors.toCollection;

// 根據id去重
         List<Person> unique = persons.stream().collect(
                collectingAndThen(
                        toCollection(() -> new TreeSet<>(comparingLong(Person::getId))), ArrayList::new)
        );

這段炫酷的程式碼是google的，還不明白是怎麼個原理，等我好好研究一下，再專門寫篇文章好好闡述一下。

還有一種寫法：

    public static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
        Map<Object, Boolean> map = new ConcurrentHashMap<>();
        return t -> map.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
    }

        // remove duplicate
        persons.stream().filter(distinctByKey(p -> p.getId())).forEach(p -> System.out.println(p));

具體解析：https://blog.csdn.net/weixin_41888813/article/details/83750984

參考: 

http://www.cnblogs.com/jizha/p/java_arraylist_duplicate.html

http://stackoverflow.com/questions/29670116/remove-duplicates-from-a-list-of-objects-based-on-property-in-java-8

來源於：

https://blog.csdn.net/jiaobuchong/article/details/54412094